What is the integral from 1 to 2 of 13sqrt(4x^2-3) ?

The integral from 1 to 2 of 13sqrt(4x^2-3) is (13(3ln(3)+4sqrt(13)-3ln(4+sqrt(13))-2))/4

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integral from 1 to 2 of 13sqrt(4x^2-3)

\int_{1}^{2} 13 4 x^{2} - 3 d x

\frac{13 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

Solution steps

\int_{1}^{2} 13 4 x^{2} - 3 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 13 \cdot \int_{1}^{2} 4 x^{2} - 3 d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} \frac{3}{2} tan^{2} (u) sec (u) d u

= 13 \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} \frac{3}{2} tan^{2} (u) sec (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 13 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 13 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 13 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 13 \cdot \frac{3}{2} (- \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (u) d u + \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (u) d u)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (u) d u = ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (u) d u = \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))

= 13 \cdot \frac{3}{2} (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)))

Simplify

13 \cdot \frac{3}{2} (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))) : \frac{13 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

= \frac{13 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

\frac{d}{d x} (- 5 x + 2)

\frac{\partial}{\partial y} (2 x - y^{2})

- \frac{5}{4 y ^{3}}

\frac{d}{d x} (cos (x^{- 5}))

\int (2 sin (θ) - csc (θ))^{2} d θ

What is the integral from 1 to 2 of 13sqrt(4x^2-3) ?
The integral from 1 to 2 of 13sqrt(4x^2-3) is (13(3ln(3)+4sqrt(13)-3ln(4+sqrt(13))-2))/4